# Spectral decomposition and invertibility

• December 3rd 2009, 07:12 PM
Last_Singularity
Spectral decomposition and invertibility
Question: Let $T$ be a normal operator on a finite-dimensional complex inner product space $V$. Consider spectral decomposition $T = \lambda_1 T_1 + ... + \lambda_k T_k$ and prove that $T$ is invertible if and only if $\lambda_i \neq 0$ for $i \leq i \leq k$.

Any pointers would be helpful - I am unable to begin currently.
• December 3rd 2009, 07:19 PM
tonio
Quote:

Originally Posted by Last_Singularity
Question: Let $T$ be a normal operator on a finite-dimensional complex inner product space $V$. Consider spectral decomposition $T = \lambda_1 T_1 + ... + \lambda_k T_k$ and prove that $T$ is invertible if and only if $\lambda_i \neq 0$ for $i \leq i \leq k$.

Any pointers would be helpful - I am unable to begin currently.

Well, this is ALWAYS true: any operator (of a finite-dimensional vector space $V$) over any field is invertible iff all its eigenvalues are different from zero, and the proof is painfully simple: $T$ is NOT invertible iff $Ker(T)\ne {0}\,\Longleftrightarrow\,\exists\,0\ne v\in V\,\,\,s.t.\,\,Tv=0=0\cdot v\,\Longleftrightarrow\,0$ is an eigenvaue of $T$.

Tonio